9
$\begingroup$

Given a symmetric monoidal category $C$, we can construct its endomorphism operad (or multicategory) $End(C)$ whose objects are the objects of $C$, and for which the multimorphisms from $\{c_1,\ldots,c_n\}$ to $c$ are given by the hom set $Hom_C(c_1\otimes\cdots c_n,c)$. From this symmetric operad we can then produce a category of operators whose objects are finite lists of objects of $C$ and whose morphisms are described (roughly) in the following way: given two lists $\{c_1,\ldots,c_n\}$ and $\{d_1,\ldots,d_m\}$, a morphism between them is a morphism $\phi:\langle n\rangle\to\langle m\rangle$ in finite pointed sets (the category often denoted $\Gamma$ by Segal and others, and $\mathcal{F}in_\ast$ by Lurie) and for each element $i\in\langle m\rangle$ and set $\{j\in\phi^{-1}(i)\}$ a multimorphism of $End(C)$ going from $\{c_j\}$ to $d_i$. Let's denote the application of these procedures to $C$ by $End(C)^\otimes$. Essentially by construction, $End(C)^\otimes$ admits a Grothendieck opfibration to $\Gamma$, and as such, corresponds to a pseudofunctor $F_C:\Gamma\to Cat$. One might even say that this pseudofunctor defines the symmetric monoidal structure on $C$.

Given a symmetric monoidal category there is a symmetric monoidal structure on $C^{op}$, so we may also consider $End(C^{op})^\otimes$. Similarly, we obtain a pseudofunctor $F_{C^{op}}:\Gamma\to Cat$.

The sort of theorem I'm after is the following:

The composition of pseudofunctors $\Gamma\overset{F_{C}}\to Cat\overset{op}\to Cat$ is equivalent to the pseudofunctor $\Gamma\overset{F_{C^{op}}}\to Cat$.

Or at the very least, I just want to say that if I apply the Grothendieck construction $\Gamma\overset{F_{C}}\to Cat\overset{op}\to Cat$ I get a symmetric monoidal category which is equivalent, as a symmetric monoidal category, to $C^{op}$ with the opposite symmetric monoidal structure.

$\endgroup$
  • 4
    $\begingroup$ Hi Jon, I'd say that both pseudo-functors are actually equal. The opposite of a product of categories is the product of opposite categories, and $F_C$ is given by $F_C(\langle n\rangle)=C^n$. $\endgroup$ – Fernando Muro Jun 14 '16 at 18:36
1
$\begingroup$

This question is answered in the affirmative in a recent paper of mine with Liang Ze Wong. In fact, we prove it more generally for a (strictly) monoidal simplicially enriched category. As any simplicially enriched monoidal category can be rigidified up to monoidal equivalence to a strict one, there's nothing lost by making the strict assumption. The enriched case of course implies the unenriched statement above. The paper is here: https://arxiv.org/abs/1808.08020

$\endgroup$
  • 1
    $\begingroup$ So what was wrong with Fernando's argument (simply unpacking the definitions of $\mathrm{op} \circ F_C$ and $F_{C^{\mathrm{op}}}$ and observing that they coincide)? $\endgroup$ – Najib Idrissi May 23 at 9:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.